(Solution Manual) Physics

Cutnell,Johnson, 9

Physics

Solution Manual

9th Edition

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INSTRUCTOR SOLUTIONS MANUAL Volume I Chapters 1 − 17 INSTRUCTOR’S SOLUTIONS MANUAL Physics Ninth Edition John D. Cutnell Kenneth W. Johnson Southern Illinois University at Carbondale SSM PREFACE for Volume I This volume contains the answers to the Focus on Concepts questions and the complete solutions to the problems for chapters 1 through 17. These chapters include all of mechanics (including fluids), thermal physics and wave motion. The solutions for chapters 18 through 32 (electricity and magnetism, light and optics, and modern phy sics) are contained in Volume 2. Each chapter is organized so that the answers to the Focus on Concepts questions appear first, followed by the solutions to the problems. An electronic version of this manual is available on the Instructor’s Companion Web site. The files are available in three formats: Microsoft Word 97-2003, Microsoft Word 2007, and PDF files of each individual solution. The icon at the beginning of some of the problems indicates that the solution is also available in the Student Solutions Manual. Note to adopters regarding The Instructor’s Solutions Manual to accompany Physics, 9e by Cutnell & Johnson Thank you for adopting Physics , 9e. We are pleased to be able to provide you with a variety of support material to help y ou in the teaching of your course. All of this material −including the Instructor’s Solutions Manual −is copyrighted by John Wiley & Sons, Inc. and is explicitly intended for use only at your institution. Please note that our providing these solutions does not carry with it permission to distribute them beyond your institution. Before putting any of the solutions on a Web site, we ask that you request formal permission from us to do so. Please write to: Permissions Department, John Wiley & Sons, Inc. 111 River St. Hoboken, NJ 07030. In most cases we will grant such permission PROVIDED THAT THE WEB SITE IS PASSWORD PROTECTED. Our goal is to prevent students from other campuses from being able to access your solutions. We trust that you can understand how that might undermine the efforts of your colleagues at other institutions. We appreciate your support and understanding in this matter. CHAPTER 1 Answers to Focus on Concepts Questions Problems 1 3 CHAPTER 2 Answers to Focus on Concepts Questions Problems 42 45 CHAPTER 3 Answers to Focus on Concepts Questions Problems 97 99 CHAPTER 4 Answers to Focus on Concepts Questions Problems 158 161 CHAPTER 5 Answers to Focus on Concepts Questions Problems 236 238 CHAPTER 6 Answers to Focus on Concepts Questions Problems 273 276 CHAPTER 7 Answers to Focus on Concepts Questions Problems 332 334 CHAPTER 8 Answers to Focus on Concepts Questions Problems 378 380 CHAPTER 9 Answers to Focus on Concepts Questions Problems 419 421 CHAPTER 10 Answers to Focus on Concepts Questions Problems 476 479 CHAPTER 11 Answers to Focus on Concepts Questions Problems 539 542 CHAPTER 12 Answers to Focus on Concepts Questions Problems 601 603 CHAPTER 13 Answers to Focus on Concepts Questions Problems 658 660 CHAPTER 14 Answers to Focus on Concepts Questions Problems 689 691 CHAPTER 15 Answers to Focus on Concepts Questions Problems 729 731 CHAPTER 16 Answers to Focus on Concepts Questions Problems 792 794 CHAPTER 17 Answers to Focus on Concepts Questions Problems 856 858 CONTENTS 4 A2 + 4 A2 x y  3.0 km  CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the last vector. 2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to the shortest distance between the tail of A and the head of B. Thus, R is less than the magnitude (length) of A plus the magnitude of B. 3. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides are known, so the Pythagorean theorem can be used to determine the length R of the hypotenuse. 4. (b) The angle is found by using the inverse tangent function,  = tan−1  4.0 km  = 53 .   5. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are not reversed. 6. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector A is not reversed. 7. (e) These vectors form a closed four -sided polygon, with the head of the fourth vector exactly meeting the tail of the first vector. Thus, the resultant vector is zero. 8. (c) When the two vector components Ax and Ay are added by the tail -to-head method, the sum equals the vector A. Therefore, these vector components are the correct ones. 9. (b) The three vectors form a right triangle, so the magnitude of A is given by the Pythagorean theorem as A = . If Ax and Ay double in size, then the magnitude of A doubles: = = 2 = 2 A. 10. (a) The angle  is determined by the inverse tangent function,  = tan−1  Ay  . If A and  A  x  x  Ay both become twice as large, the ratio does not change, and  remains the same. 11. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar component along the x axis ( Ax = 0 m) and its scalar component along the y axis is negative. A2 + A2 x y ( 2 A + 2 A ) 2 ( y ) 2 x A2 + A2 x y 2 INTRODUCTION AND MATHEMATICAL CONCEPTS 12. (e) The scalar components are given by Ax = −(450 m) sin 35.0  = −258 m and Ay = −(450 m) cos 35.0  = −369 m. 13. (d) The distance (magnitude) traveled by each runner is the same, but the directions are different. Therefore, the two displacement vectors are not equal. 14. (c) Ax and Bx point in opposite directions, and Ay and By point in the same direction. 15. (d) 16. Ay = 3.4 m, By = 3.4 m 17. Rx = 0 m, Ry = 6.8 m 18. R = 7.9 m,  =  degrees Chapter 1 Problems 3 ft ft 1 mi mi h 1 mi SSM CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS PROBLEMS 1. REASONING We use the fact that 1 m = 3.28 ft to form the following conversion factor: (1 m)/(3.28 ft) = 1. SOLUTION To convert ft2 into m2, we apply the conversion factor twice: Area = (1330 ) 1 m  1 m  =  3.28  3.28     2. REASONING a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we need to convert miles to kilometers. This conversion is achieved by using the relation 1.609 km = 1 mi (see the page facing the inside of the front cover of the text). b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must convert miles to meters and hours to seconds. This is accomplished by using the conversions 1 mi = 1609 m and 1 h = 3600 s. SOLUTION a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi) = 1, we find the speed of the bicyclists is  mi   mi  1.609 km  Speed =  34.0 h (1) =  34.0 h   =      b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and (1 h)/(3600 s) = 1, the speed of the bicyclists is  mi    1609 m  1h  Speed =  34.0 h (1)(1) =  34.0   3600s  =       3. REASONING We use the facts that 1 mi = 5280 ft, 1 m = 3.281 ft, and 1 yd = 3 ft. With these facts we construct three conversion factors: (5280 ft)/(1 mi) = 1, (1 m)/(3.281 ft) = 1, and (3 ft)/(1 yd) = 1. ft2 15.2 m s 54.7 km h 124 m2 4 INTRODUCTION AND MATHEMATICAL CONCEPTS SOLUTION By multiplying by the given distance d of the fall by the appropriate conversion factors we find that  5280 ft  1 m   3 ft  1 m  d = (6 mi ) 1 mi  3.281 ft  + (551 yd ) 1 yd  3.281 ft  =       4. REASONING The word “per” indicates a ratio, so “0.35 mm per day” means 0.35 mm/d, which is to be expressed as a rate in ft/century. These units differ from the given units in both length and time dimensions, so both must be converted. For length, 1 m = 103 mm, and 1 ft = 0.3048 m. For time, 1 year = 365.24 days, and 1 century = 100 years. Multiplying the resulting growth rate by one century gives an estimate of the total length of hair a long -lived adult could grow over his lifetime. SOLUTION Multiply the given growth rate by the length and time conversion factors, making sure units cancel properly: Growth rate =  0.35 mm  1 m  1 ft  365.24 d  100 y  =  d  103 mm  0.3048 m  1y  century        5. REASONING In order to calculate d, the units of a and b must be, respectively, cubed and squared along with their numerical values, then combined algebraically with each other and the units of c. Ignoring the values and working first with the units alone, we have a3 (m)3 m 3 2 m2 d = cb2 → (m/s)(s)2 = ( m/s )s 21 = s Therefore, the units of d are m2/s. SOLUTION With the units known, the numerical value may be calculated: (9.7)3 2 d = (69)(4.2)2 m /s = 6. REASONING The dimensions of the variables v, x, and t are known, and the numerical factor 3 is dimensionless. Therefore, we can solve the equation for z and then substitute the known dimensions. The dimensions L and T can be treated as algebraic quantities to determine the dimensions of the variable z. 0.75 m2/s 42 ft/century 10 159 m Chapter 1 Problems 5 1.5 liters SSM ( ) SOLUTION Since v = 1 zxt2 , it follows that 3 z = 3v xt2 . We know the following dimensions: v = L / T, of x = L, and t = T. Since the factor 3 is dimensionless, z has the dimensions v = L / T = xt2 LT2 7. REASONING This problem involves using unit conversions to determine the number of magnums in one jeroboam. The necessary relationships are 1.0 magnum = 1.5 liters 1.0 jeroboam = 0.792 U. S. gallons 1.00 U. S. gallon = 3.785 10–3 m3 = 3.785 liters These relationships may be used to construct the appropriate conversion factors. SOLUTION By multiplying one jeroboam by the appropriate conversion factors we can determine the number of magnums in a jeroboam as shown below:  0.792 gallons  3.785 liters  1.0 magnum  (1.0 jeroboam )  1.0 jeroboam  1.0 gallon   =     8. REASONING By multiplying the quantity 1.78  10−3 kg/ (sm) by the appropriate conversions factors, we can convert the quantity to units of poise (P). These conversion factors are obtainable from the following relationships between the various units: 1 kg = 1.00 1 m = 1.00  103 g  102 cm 1 P = 1 g/ (s cm) SOLUTION The conversion from the unit kg/ (s m) to the unit P proceeds as follows:  −3   1 P  1.78  10     =  1 g / s  cm  kg s  m  1.00  103 g     1 kg    1 m 1.00  102 cm 1.78  10−2 P 2.0 magnums 1 T3 6 INTRODUCTION AND MATHEMATICAL CONCEPTS  1 gal  128 oz      3.785 10−3 1 gal m3   1 mL   10 m  −6 3   U.S. gallons   9. REASONING Multiplying an equation by a factor of 1 does not alter the equation; this is the basis of our solution. We will use factors of 1 in the following forms: 1 gal 128 oz = 1, since 1 gal = 128 oz 3.785 10−3 m3 = 1, since 3.785  10−3 m3 = 1 gal 1 gal 1 mL 10−6 m3 = 1, since 1 mL = 10−6 m3 SOLUTION The starting point for our solution is the fact that Volume = 1 oz Multiplying this equation on the right by factors of 1 does not alter the equation, so it follows that Volume = (1 oz)(1)(1)(1) = (1 oz ) = Note that all the units on the right, except one, are eliminated algebraically, leaving only the desired units of milliliters (mL). 10. REASONING To convert from gallons to cubic meters, use the equivalence 1 U.S. gal = 3.785×10−3 m3. To find the thickness of the painted layer, we use the fact that the paint’s volume is the sam e, whether in the can or painted on the wall. The layer of paint on the wall can be thought of as a very thin “box” with a volume given by the product of the surface area (the “box top”) and the thickness of the layer. Therefore, its thickness is the ratio of the volume to the painted surface area: Thickness = Volume/Area . That is, the larger the area it’s spread over, the thinner the layer of paint. SOLUTION a. The conversion is (0.67 U.S. gallons ) 3.785 10−3 m3  =   b. The thickness is the volume found in (a) divided by the area, Thickness = Volume = 2.5 10−3 m3 = Area 13 m2 1.9 10−4 m 2.5 10−3 m3 29.6 mL Chapter 1 Problems 7 5.5 km h2 + h2 o a (35.0 km) 2 + (72.0 km) 2 SSM SSM   11. REASONING The dimension of the spring constant k can be determined by first solving the equation T = 2 for k in terms of the time T and the mass m. Then, the dimensions of T and m can be substituted into this expression to yield the dimension of k. SOLUTION Algebraically solving the expression above for k gives k = 4 2m / T 2 . The term 4 2 is a numerical factor that does not have a dimension, so it can be ignored in this analysis. Since the dimension for mass is [M] and that for time is [T], the dimension of k is Dimension of k = 12. REASONING AND SOLUTION The following figure (not drawn to scale) shows the geometry of the situation, when the observer is a distance r from the base of the arch. The angle  is related to r and h by tan = h/ r. Solving for r, we find r = h = 192 m = 5.5 103 m = h = 192 m tan tan 2.0 r 13. 1 3 . REASONING The shortest distance between the two towns is along the line that joins them. This distance, h, is the hypotenuse of a right triangle whose other sides are ho = 35.0 km and ha = 72.0 km, as shown in the figure below. SOLUTION The angle  is given by tan = ho / ha so that  = tan−1  35.0 km = 72.0 km We can then use the Pythagorean theorem to find h. h = = = m / k  80.1 km 25.9 S of W M T2 8 INTRODUCTION AND MATHEMATICAL CONCEPTS L2 + L2 2 0.35 m 2  14. REASONING The drawing shows a schematic representation of the hill. We know that the hill rises 12.0 m vertically for every 100.0 m of distance in the h horizontal direction, so that ho = 12.0 m and o ha = 100.0 m. Moreover, according to Equation 1.3, the tangent function is tan = ho / ha . Thus, we can use the h inverse tangent function to determine the angle θ. a SOLUTION With the aid of the inverse tangent function (see Equation 1.6) we find that  = tan−1  ho  = tan−1  12.0 m  =  h   100.0 m   a    15. REASONING Using the Pythagorean theorem (Equation 1.7), we find that the relation between the length D of the diagonal of the square (which is also the diameter of the circle) and the length L of one side of the square is D = = 2L . SOLUTION Using the above relation, we have D = 2L or L = D = = 16. REASONING In both parts of the drawing the line of sight, the horizontal dashed line, and the vertical form a right triangle. The angles a = 35.0° and b = 38.0° at which the person’s line of sight rises above the horizontal are known, as is the horizontal distance d = 85.0 m from the building. The unknown vertical sides of the right triangles correspond, respectively, to the heights Ha and Hb of the bottom and top of the antenna relative to the person’s eyes. The antenna’s height H is the difference between Hb and Ha: H = Hb − Ha . The horizontal side d of the triangle is adjacent to the angles a and b, while the vertical sides Ha and Hb are opposite these angles. Thus, in either triangle, the angle  is related to the horizontal and vertical sides by Equation 1.3  tan = ho  :  h   a  tan = Ha (1) a d tan = Hb (2) b d 0.25 m 6.84 Chapter 1 Problems 9 (a) (b) SOLUTION Solving Equations (1) and (2) for the heights of the bottom and top of the antenna relative to the person’s eyes, we find that Ha = d tana and Hb = d tanb The height of the antenna is the difference between these two values: H = Hb − Ha = d tanb − d tana = d (tanb − tana ) H = (85.0 m)(tan 38.0 17. REASONING The drawing shows the heights of the two balloonists and the horizontal distance x between them. Also shown in dashed lines is a right triangle, one angle of which is 13.3. Note that the side adjacent to the 13.3 13.3 angle is the horizontal distance x, while the side opposite the angle is the distance between the two heights, 61.0 m − 48.2 m. Since we know the angle and the length of one side of the right triangle, we can use trigonometry to find the length of the other side. x 48.2 m 61.0 m SOLUTION The definition of the tangent function, Equation 1.3, can be used to find the horizontal distance x, since the angle and the length of the opposite side are known: Solving for x gives tan13.3 = length of opposite side length of adjacent side (= x) x = length of opposite side = 61.0 m − 48.2 m =  tan13.3  tan13.3  H Ha Hb a b d d − tan 35.0 ) = 6.9 m 54.1 m 10 INTRODUCTION AND MATHEMATICAL CONCEPTS 99 51 30 18. REASONING As given in Appendix E, the law of cosines is c2 = a2 + b2 − 2ab cos c a γ where c is the side opposite angle γ, and a and b are the other b two sides. Solving for γ, we have  = cos−1  a2 + b2 − c2   2ab    SOLUTION For c = 95 cm, a = 150 cm, and b = 190 cm −  a2 + b2 − c2  (150 cm)2 + (190)2 −(95 cm)2   = cos 1   = cos−1   = 30  2ab   2(150 cm)(190 cm)  Thus, the angle opposite the side of length 95 cm is . Similarly, for c = 150 cm, a = 95 cm, and b =190 cm, we find that the angle opposite the side of length 150 cm is . Finally, for c = 190 cm, a = 150 cm, and b = 95 cm, we find that the angle opposite the side of length 190 cm is . As a check on these calculations, we note that 30+ 51+ 99 =180, which must be the case for the sum of the three angles in a triangle. 19. REASONING Note from the drawing that the shaded right triangle contains the angle  , the side opposite the angle (length = 0.281 nm), and the side adjacent to the angle (length = L). If the length L can be determined, we can use trigonometry to find . The bottom face of the cube is a square whose diagonal has a length L. This length can be found from the Pythagorean theorem, since the lengths of the two sides of the square are known. SOLUTION The angle can be obtained from the  0.281 nm L 0.281 nm 0.281 nm inverse tangent function, Equation 1.6, as  = tan−1 (0.281 nm) / L . Since L is the length of the hypotenuse of a right triangle whose sides have lengths of 0.281 nm, its value can be determined from the Pythagorean theorem: Chapter 1 Problems 11 (0.281 nm)2 + (0.281 nm)2 (R + h)2 − R2 2Rh + h2 2(6.38106 m)(1.6 m) + (1.6 m)2 m L = = 0.397 nm Thus, the angle is  = tan−1  0.281 nm  = tan−1  0.281 nm  =  L   0.397 nm      20. REASONING a. The drawing shows the person standing on the earth and looking at the horizon. Notice the right triangle, the sides of which are R, the radius of the earth, and d, the distance from the person’s eyes to the horizon. The length of the hypotenuse is R + h, where h is the height of the person’s eyes above the water. Since we know the lengths of two sides of the triangle, the Pythagorean theorem can be employed to find the length of the third side. Horizon b. To convert the distance from meters to miles, we use the relation 1609 m = 1 mi (see the page facing the inside of the front cover of the text). SOLUTION a. The Pythagorean theorem (Equation 1.7) states that the square of the hypotenuse is equal to the sum of the squares of the sides, or (R + h)2 = d 2 + R2 . Solving this equation for d yields d = = = = = b. Multiplying the distance of 4500 m by a factor of unity, (1 mi)/(1609 m) = 1, the distance (in miles) from the person's eyes to the horizon is d = (4500 m )(1) = (4500 m ) 1 mi  =  1609    R2 + 2Rh + h2 − R2 h d R 90 R Earth 2.8 mi 4500 m 35.3 12 INTRODUCTION AND MATHEMATICAL CONCEPTS 1.2 102 m 21. 2 1 . shows REASONING The drawing at the right the location of each deer A, B, and C. From the problem statement it follows that b = 62 m c = 95 m  = 180 − 5 1 − 7 7 = 5 2 Applying the law of cosines (given in Appendix E) to the geometry in the figure, we have a 2 − 2 ab cos  + ( b2 − c2 ) = 0 which is an expression that is quadratic in a. It can be simplified to Aa2 + Ba + C = 0 , with A = 1 B= –2b cos  = – 2(62 m) cos 52 = –76 m C = ( b2 − c2 ) = (62 m) 2 − (95 m)2 = –5181 m2 This quadratic equation can be solved for the desired quantity a. SOLUTION Suppressing units, we obtain from the quadratic formula a = 2(1) = 1.2  102 m and – 43 m Discarding the negative root, which has no physical significance, we conclude that the distance between deer A and C is . 22. REASONING The trapeze cord is L = 8.0 m long, so that the trapeze is initially h1 = L cos 41° meters below the support. At the instant he releases the trapeze, it is h2 = L cos  meters below the h2 support. The difference in the heights is d = h2 – h1 = 0.75 m. Given that the   41° L L h1 41° trapeze is released at a lower elevation than the platform, we expect to find  < 41°. SOLUTION Putting the above relationships together, we have 0.75 m −( −76)  ( −76) 2 − 4(1)( –5181) SSM Chapter 1 Problems 13 A2 + b− B g2 b445 Ng2 + b−325 Ng2 SSM  L H K44 5 N H K44 5 N d = h2 − h1 = L cos − L cos 41 cos = d + cos 41 L  = cos−1  d + cos 41  23. REASONING a. Since the two force vectors A and B have directions due west and due north, they are perpendicular. Therefore, the resultant vector F = A + B has a magnitude given by the Pythagorean theorem: F2 = A2 + B2. Knowing the magnitudes of A and B, we can calculate the magnitude of F. The direction of the resultant can be obtained using trigonometry. b. For the vector F = A – B we note that the subtraction can be regarded as an addition in the following sense: F = A + (–B). The vector –B points due south, opposite the vector B, so the two vectors are once again perpendicular and the magnitude of F again is given by the Pythagor ean theorem. The direction again can be obtained using trigonometry. SOLUTION a. The drawing shows the two vectors and the resultant vector. According to the Pythagorean theorem, we have F 2 = A 2 + B 2 B F = F = = North (a) North (b) Using trigonometry, we can see that the direction of the resultant is tan = B A or  = tan−1 FG 325 N IJ = b. Referring to the drawing and following the same procedure as in part a, we find F  2 = A2 + b− B g2 or F  = = = tan = B A or  = tan−1 FG 325 N IJ =  F –B A F  A    = cos −1   8.0 m 0.75 m  + cos 41 = 32    or d + L cos 41 = L cos A 2 + B 2 b445 Ng2 + b325 Ng2 36.1 south of west 551 N 36.1 north of west 551 N 14 INTRODUCTION AND MATHEMATICAL CONCEPTS F3 + F4 = 70.0 newtons, due west F1 + F3 = 10.0 newtons, due east SSM due west due east 24. REASONING Since the initial force and the resultant force point along the east/west line, the second force must also point along the east/west line. The direction of the second force is not specified; it could point either due east or due west, so there are two answers. We use “N” to denote the units of the forces, which are specified in newtons. SOLUTION If the second force points , both forces point in the same direction and the magnitude of the resultant force is the sum of the two magnitudes: F1 + F2 = FR. Therefore, F2 = FR – F1 = 400 N – 200 N = If the second force points magnitude of the resultant force is Therefore, , the two forces point in opposite directions, and the the difference of the two magnitudes: F2 – F1 = FR. F2 = FR + F1 = 400 N + 200 N = 25. REASONING For convenience, we can assign due east to be the positive direction and due west to be the negative direction. Since all the vectors point along the same east - west line, the vectors can be added just like the usual algebraic addition of positive and negative scalars. We will carry out the addition for a ll of the possible choices for the two vectors and identify the resultants with the smallest and largest magnitudes. SOLUTION There are six possible choices for the two vectors, leading to the following resultant vectors: F1 + F2 = 50.0 newtons +10.0 newtons = +60.0 newtons = 60.0 newtons, due east F1 + F3 = 50.0 newtons − 40.0 newtons = +10.0 newtons = 10.0 newtons, due east F1 + F4 = 50.0 newtons − 30.0 newtons = +20.0 newtons = 20.0 newtons, due east F2 + F3 = 10.0 newtons − 40.0 newtons = −30.0 newtons = 30.0 newtons, due west F2 + F4 = 10.0 newtons − 30.0 newtons = −20.0 newtons = 20.0 newtons, due west F3 + F4 = −40.0 newtons − 30.0 newtons = −70.0 newtons = 70.0 newtons, due west The resultant vector with the smallest magnitude is . The resultant vector with the largest magnitude is . 600 N 200 N Chapter 1 Problems 15 (63 units)2 + (63 units)2 (63 units)2 + (63 units)2 (1080 m)2 + (520 m)2  63 units   63 units  26. REASONING The Pythagorean theorem (Equation 1.7) can be used to find the magnitude of the resultant vector, and trigonometry can be employed to determine its direction. a. Arranging the vectors in tail -to-head fashion, we can see that the vector A gives the resultant a westerly direction and vector B gives the resultant a southerly direction. Therefore, the resultant A + B points south of west. b. Arranging the vectors in tail -to-head fashion, we can see that the vector A gives the resultant a westerly direction and vector –B gives the resultant a northerly direction. Therefore, the resultant A + (–B) points north of west. SOLUTION Using the Pythagorean theorem and trigonometry, we obtain the following results: a. Magnitude of A + B = =  = tan−1  63 units  =   b. Magnitude of A − B = =  = tan−1  63 units  =   27. REASONING At the turning point, the distance to the campground is labeled d in the drawing. Note that d is the length of the hypotenuse of a right triangle. Since we know the lengths of the other two sides of the triangle, the Pythagorean theorem can be used to find d. Turning N point  d  Campground The direction that cyclist #2 must head during the last part of the trip is given by the angle . It can be determined by using the inverse tangent function. SOLUTION a. The two sides of the triangle have lengths of 1080 m and 520 m (1950 m − 1430 m = 520 m). The length d 1950 m Start W S 1080 m 1430 m E of the hypotenuse can be determined from the Pythagorean theorem, Equation (1.7), as d = = 1200 m 45 north of west 89 units 45 south of west 89 units 16 INTRODUCTION AND MATHEMATICAL CONCEPTS s2 + s2 2s2 2 J 2 + K 2 16d 2 + 4d 2 20d 2 40 40 K D J d s s  1080 m  b. Since the lengths of the sides opposite and adjacent to the angle  are known, the inverse tangent function (Equation 1.6) can be used to find .  = tan−1  520 m  =   28. REASONING The triple jump consists of a double jump in one direction, followed by a perpendicular single jump, which we can represent with displacement vectors J and K (see the drawing). These two perpendicular vectors form a right triangle with their resultant D = J + K, which is the displacement of the colored checker. In order to find the magnitude D of the displacement, we first need to find the magnitudes J and K of the double jump and the single jump. As the three sides of a right triangle, J, K, and D (the hypotenuse) are related to one another by the Pythagorean theorem (Equation 1.7) The double jump moves the colored checker a straight -line distance equal to the length of four square’s diagonals d, and the single jump moves a length equal to two square’s diagonals. Therefore, J = 4d and K = 2d (1) Let the length of a square’s side be s. Any two adjacent sides of a square form a right triangle with the square’s diagonal (see the drawing). The Pythagorean theorem gives the diagonal length d in terms of the side length s: d = = = s (2) SOLUTION First, we apply the Pythagorean theorem to the right triangle formed by the three displacement vectors, using Equations (1) for J and K: D = = = = = d (3) Substituting Equation (2) into Equation (3) gives D = d = (s 2 ) = s = (4.0 cm) = (4d )2 + (2d )2 20 20 20 25 cm 26 south of east Chapter 1 Problems 17 (44.0 cm)2 + (12.0 cm)2 (16.0 cm)2 + (36.0 cm)2 29. REASONING Both P and Q and the vector sums K and M can be drawn with P correct magnitudes and directions by Q counting grid squares. To add vectors, place them tail-to-head and draw the M resultant vector from the tail of the first vector to the head of the last. The vector 2P is equivalent to P + P, and −Q is a −Q vector that has the same magnitude as Q, except it is directed in the opposite K direction. P The vector M runs 11 squares P horizontally and 3 squares vertically, and the vector K runs 4 squares horizontally 8.00 cm and 9 squares vertically. These distances can be converted from grid squares to centimeters with the grid scale: 1 square = 4.00 cm. Once the distances are calculated in centimeters, the Pythagorean theorem (Equation 1.7) will give the magnitudes of the vectors. SOLUTION a. The vector M = P + Q runs 11 squares horizontally and 3 squares vertically, and these distances are equivalent to, respectively,  4.00 cm (11 squares ) = 44.0 cm and  square     4.00 cm (3 squares ) = 12.0 cm . Thus, the magnitude of M is  square    M = = b. Similarly, the lengths of the horizontal and vertical distances of K = 2P − Q are 4 horizontal squares and 9 vertical squares, or 16.0 cm and 36.0 cm, respectively. The magnitude of K is then K = = 30. REASONING a. and b. The following drawing shows the two vectors A and B, as well as the resultant vector A + B. The three vectors form a right triangle, of which two of the sides are known. We can employ the Pythagorean theorem, Equation 1.7, to find the length of the third side. The angle  in the drawing can be determined by using the inverse cosine function, Equation 1.5, since the side adjacent to  and the length of the hypotenuse are known. 39.4 cm 45.6 cm 18 INTRODUCTION AND MATHEMATICAL CONCEPTS R2 − A2 (15.0 units)2 −(12.3 units)2 12.3 units  units A + B A B 15.0 12.3 units −B 15.0 units  A A − B SSM 34.9 south of west  15.0 units  N W E S c. and d. The following drawing shows the vectors A and −B, as well as the resultant vector A − B. The three vectors form a right triangle, which is identical to the previous drawing, except for orientation. Thus, the lengths of the hypotenuses and the angles are equal. N W E S SOLUTION a. Let R = A + B. The Pythagorean theorem (Equation 1.7) states that the square of the hypotenuse is equal to the sum of the squares of the sides, so that B yields R2 = A2 + B2 . Solving for B = = = b. The angle  can be found from the inverse cosine function, Equation 1.5:  = cos−1  12.3 units  =   c. Except for orientation, the triangles in the two drawings are the same. Thus, the value for B is the same as that determined in part (a) above: B = d. The angle  is the same as that found in part (a), except the resultant vector points south of west, rather than north of west:  = 31. REASONING AND SOLUTION The single force needed to produce the same effect is equal to the resultant of the forces provided by the two ropes. The following figure shows the force vectors drawn to scale and arranged tail to head. The magnitude and direction of the resultant can be found by direct measurement using th e scale factor shown in the figure. 8.6 units 34.9 north of west 8.6 units Chapter 1 Problems 19 (3.75 km)2 − (2.50 km)2 5600 N H K2.5 0 km H K2.50km a. From the figure, the magnitude of the resultant is . b. The single rope should be directed along the dashed line in the text drawing. 32. REASONING a. Since the two displacement vectors A and B have directions due south and due east, they are perpendicular. Therefore, the resultant vector R = A + B has a magnitude given by the Pythagorean theorem: R2 = A2 + B2. Knowing the magnitudes of R and A, we can c alculate the magnitude of B. The direction of the resultant can be obtained using trigonometry. b. For the vector R = A – B we note that the subtraction can be regarded as an addition in the following sense: R = A + (–B). The vector –B points due west , opposite the vector B, so the two vectors are once again perpendicular and the magnitude of R again is given by the Pythagorean theorem. The direction again can be obtained using trigonometry. SOLUTION a. The drawing shows the two vectors and the resultant vector. According to the Pythagorean theorem, we have R2 = A2 + B2 or B = B = = (a) (b) Using trigonometry, we can see that the direction of the resultant is tan = B A or  = tan−1 FG 2.8 km IJ = b. Referring to the drawing and following the same procedure as in part a, we find R  2 = A2 + b− B g2 or B = = b3.75 kmg2 − b2.50 kmg2 = tan = B A or  = tan−1 FG 2.8 km IJ = R2 − A2 R  2 − A2 North A  R B North R  A –B 48 west of south 2.8 km 48 east of south 2.8 km 20 INTRODUCTION AND MATHEMATICAL CONCEPTS SSM 78 N 33. REASONING AND SOLUTION The following figure is a scale diagram of the forces drawn tail -to-head. The scale factor is shown in the figure. The head of F3 touches the tail of F1, because the resultant of the t hree forces is zero. a. From the figure, F3 must have a  60.0° magnitude of if the resultant force acting on the ball is zero. b. Measurement with a protractor F3  60.0° F1 = 50.0 N indicates that the angle . Scale Factor: F2 = 90.0 N 20.0 N 34. REASONING The magnitude of the x-component of the force vector is the product of the magnitude of the force times the cosine of the angle between the vector and the x axis. Since the x-component points in the + x direction, it is positive. Likewise, the magnitude of the y component of the force vector is the product of the magnitude of the force times the sine of the angle between the vector and the x axis. Since the vector points 36.0º below the positive x axis, the y component of the vector points in the −y direction; thus, a minus sign must be assigned to the y-component to indicate this direction. SOLUTION The x and y scalar components are a. Fx = (575 newtons) cos 36.0 = b. Fy = –(575 newtons) sin 36.0 = 35. REASONING AND SOLUTION In order to determine which vector has the largest x and y components , we calculate the magnitude of the x and y components explicitly and compare them. In the calculations, the symbol u denotes the units of the vectors. Ax = (100.0 u) cos 90.0° = 0.00 u Ay = (100.0 u) sin 90.0° = 1.00  102 u Bx = (200.0 u) cos 60.0° = 1.00  102 u By = (200.0 u) sin 60.0° = 173 u Cx = (150.0 u) cos 0.00° = 150.0 u Cy = (150.0 u) sin 0.00° = 0.00 u a. b.  = 34 B has the largest y component. C has the largest x component. –338 newtons 465 newtons Chapter 1 Problems 21 v 34º 90º 36. REASONING The triangle in the drawing is a right triangle. We know one of its angles is 30.0, and the length of the hypotenuse is 8.6 m. Therefore, the sine and cosine functions can be used to find the magnitudes of Ax and Ay. The directions of these vectors can be found by examining the diagram. SOLUTION a. The magnitude Ax of the displacement vector Ax is related to the length of the hypotenuse and the 30.0  angle by the sine function (Equation 1.1). The drawing shows that the direction of Ax is due east. N #1 W E S 30.0 Ax = Asin 30.0 = ( m)sin 30.0 = b. In a similar manner, the magnitude Ay of Ay can be found by using the cosine function (Equation 1.2). Its direction is due south. Ay = Acos 30.0 = ( m)cos 30.0 = Ay 90 #2 8.6 m A Ax Net 37. REASONING Using trigonometry, we can determine the angle  from the relation tan  = Ay /Ax: SOLUTION a.  = tan−1  Ay  = tan−1  12 m  =  A   12 m   x    b.  = tan−1  Ay  = tan−1  12 m  =  A   17 m   x    c.  = tan−1  Ay  = tan−1  17 m  =  A   12 m   x    38. REASONING The drawing assumes that the horizontal direction along the ground is the x direction and shows the plane’s velocity vector v, along with its horizontal vy component vx and vertical component vy. These components, together with the vx velocity vector, form a right triangle, as indicated. Based on this right triangle, we 45 35 55  m, due south  m, due east